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Re: st: Standard normal Depvar


From   Austin Nichols <[email protected]>
To   [email protected]
Subject   Re: st: Standard normal Depvar
Date   Thu, 6 Aug 2009 12:37:13 -0400

Evans Jadotte<[email protected]> :
Your desideratum
"How can I transform the Depvar in order to force  xb^ to take on
positive values?"
and comments make no sense: N.B. exponentiation preserves rank.

You can make x positive by

su x
replace x=x+2*abs(r(min))

for example, but then you can't take the square root and have it make sense.

Maybe you want

g y=sign(x)*sqrt(abs(x))

or somesuch?

On Thu, Aug 6, 2009 at 12:26 PM, Evans Jadotte<[email protected]> wrote:
> Nick Cox wrote:
>>
>> Exponentiation will get you all positives. After that many options are
>> open.
>>
>> Evans Jadotte wrote:
>>
>>> Nick Cox wrote:
>>>>
>>>> This produces zero or positive values.
>>>>
>>>> Less pedantically, if the variable is already standard normal, why does
>>>> it need transforming?
>>>>
>>>> Nick
>>>>
>>>> Maarten buis wrote:
>>>>
>>>>> --- On Wed, 5/8/09, Evans Jadotte wrote:
>>>>>>
>>>>>> I am trying to run a regression where the dependent
>>>>>> variable has a standard normal distribution (those of you
>>>>>> familiar with the "wealth index based on the PCA analysis",
>>>>>> this is my Depvar).  However, I need to have the prediction  to be all
>>>>>> positive to use for transforming.
>>>>>> How can I transform  the Depvar in order  to
>>>>>> force  xb^ to take on positive values?
>>>>>
>>>>> Here is one option:
>>>>>
>>>>> reg y x1 x2
>>>>> predict yhat
>>>>> sum yhat, meanonly
>>>>> gen yhatprime = yhat + abs(r(min))
>>>>
>>>> *
>>>> *   For searches and help try:
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>>>> *   http://www.ats.ucla.edu/stat/stata/
>>>
>>> Thanks to Maarten and Nick for their insight and comment on my question.
>>>
>>> I have both negative values and zeros in the /depvar /(/y/)/, /so/ /the
>>> forecast, xb,  will reflect such values. And as I will need  sqrt(xb) for
>>> further transformation at later stages, I need to transform /y/ so that xb
>>> takes on all 'strictly' positive values and still preserve normality of /y/.
>>> Maarten's suggestion indeed generates a 0 and the transformation I need is
>>> in y (not yhat = xb). I have been trying a Box-Cox power transform but
>>> results are not satisfactory.
>>>
>>
>> *
>> *   For searches and help try:
>> *   http://www.stata.com/help.cgi?search
>> *   http://www.stata.com/support/statalist/faq
>> *   http://www.ats.ucla.edu/stat/stata/
>
> Thanks Nick. However, exponentiation will result in a re-ranking of
> individuals, which I must avoid. For instance, someone with a score -5
> compared with one whose score is 4, the former will end up being ranked
> higher than the latter after exponentiating. I need to preserve the ranks
> and normality after transforming.
>
> Thanks for the feedback,
>
> Evans
> *
> *   For searches and help try:
> *   http://www.stata.com/help.cgi?search
> *   http://www.stata.com/support/statalist/faq
> *   http://www.ats.ucla.edu/stat/stata/
>

*
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